I have a question regarding the conditional expected value of a ratio of two random variables coming from a Multinomial distribution.
Specifically, given for a multinoial distribution with N trials where each trial independently resulting in an income in one of the K categories with probability $p_1, p_2, ..., p_k$ where the sum of $p_1, ..., p_k$ is 1. Let denote $X_1$ to be the number of trials falling into category 1, and $X_2$ be number of trials results falling into category 2, ... and $x_1 + x_2 + ... + x_k = N$.
My question is to calculate $E(\frac{X_1}{X_2} | X_2 > 0)$ i.e. expected value of $X_1$ over $X_2$ given $X_2$ is bigger than zero, is there closed form solution? or the expected value of the above will be in a summation form?
My answer is this:
$ E(\frac{X_1}{X_2} | X_2 > 0) = \sum_{x_1=0}^{n}\sum_{x_2=1}^{n-x_1} \frac{x_1}{x_2} P(x_1, x_2)/ P(X_2 > 0)$.
So I have the denominator to be: $P(X_2>0)$ equal to $1-(1-p_2)^N$ and the upper part after the summation is $\frac{x_1}{x_2}$ ${N\choose x_1}$ ${N-x_1\choose x_2}p_1 ^ {x_1} p_2^{x_2}(1-p_1-p_2)^{N-x_1-x_2}$.
Overall I have this expression as the expected value:
$ E(\frac{X_1}{X_2} | X_2 > 0) = \sum_{x_1=0}^{n}\sum_{x_2=1}^{n-x_1} \frac{x_1}{x_2} \frac{P(x_1, x_2)}{P(X_2 > 0)} = \sum_{x_1=0}^{n}\sum_{x_2=1}^{n-x_1} \frac{x_1}{x_2} \frac{{N\choose x_1}{N-x_1\choose x_2}p_1 ^ {x_1} p_2^{x_2}(1-p_1-p_2)^{N-x_1-x_2}}{1-(1-p_2)^N} $
I am just wondering is this the most "simplified" expected value one can arrive at? or there could be some closed-form solution without the "summation"?